<u>Answer:</u>
Total displacement traveled = 298
<u>Explanation:</u>
According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).
So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.
Then after a little it can travel up another cm to reach the top.
Therefore, the total displacement traveled = 297 + 1 = 298 cm
Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;
Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
Therefore, the extension of the wire is 0.362 mm.
The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.
I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
Answer:
1.2 rad/s
Explanation:
m1 = 15 g, m2 = 9 g, ω1 = 0.75 rad/s
Let the new angular speed is ω2 and the radius of the table be r.
The angular momentum is conserved when no external torque is applied.
I1 ω1 = I2 ω2
(m1 + m2)x r^2 x 0.75 = m1 x r^2 x ω2
(15 + 9) x 0.75 = 15 x ω2
ω2 = 1.2 rad/s